Explanation:
- Final velocity, v = 200 m/s
- Initial velocity, u = 0 m/s [Starts from rest]
- Acceleration, a = 6 m/s²
<u>To calculate</u> : Time (t)
★ v = u + at
→ 200 = 0 + 6t
→ 200 = 6t
→ 200/6 = t
→ <u>33.34 s ≈ t</u>
Yeah You are doing it correct
Answer:
its terminal velocity is 19.70 m/s
the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s
Explanation:
Firstly,
given that
m = 580g = 0.58kg
Area A = 0.11 * 0.22 = 0.0242m
g = 9.8
idensity constant p = 1.21 kg/m^3
the terminal velocity of the sphere Vt is ;
Vt = √ ( 2mg / pCA)
we substitute
Vt = √ ( (2*0.58*9.8) / (1.21*1*0.0242)
Vt = √ (11.368 / 0.029282)
Vt = √ ( 388.22)
Vt = 19.70 m/s
its terminal velocity is 19.70 m/s
What will be the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance?
The Velocity of the person is;
V2 = √ 2ax
V2 = √ ( 2 * 9.8 * 4 )
V2 = √ (78.4)
V2 = 8.85 m/s
the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s
Electrons are added to your body as you shuffle your feet across the room. Then the shock is when they discharge from your body to the doorknob when you touch it.
I believe it’s 1 aka A part of it is reflected as Ray 2